SOLUTION SET Chapter 2 WAVE NATURE OF LIGHT

SOLUTION SET
Chapter 2
WAVE NATURE OF LIGHTTHE INTERACTION OF LIGHT WITH MATERIALS
"LASER FUNDAMENTALS"
Second Edition
By William T. Silfvast
CH 21. Show how to go from the integral form to the differential form of Maxwell's four
equations.
~ {ver~el1ce .l la.o reVV\
f E· iS
5
~
(b)
\7·
E Jv =
v
t r\7· 5
(
r~ ~)Jv
v
.lV=
i. [rohl
~ B. e1S =- \ \7. 8 .AV== D ~
s
v
S.io k e ~
lC)
J
t
J,
)
~Q.B:f
7. B::
0
lh e. b r-e.. V\i\
B·il ~ )\7x8·JA=µol"'f[µD(J+g:Q),dA
A
A
=-/ 7 x B ::: ,µD ( "f + ~)
sTo Ile~
~or~~
tJft:Ja
EI e/-e
=
.
v
xE . d.A
=
f
A
d-:r
ffA
- JQ
- B. dA-
~
{ H 2.
2. Calculate the electrical force of attraction between a positive and negative charge
separated by a distance of 5 .3 x io- 11 rn (the average distance between an electron
in its ground state and the nucleus). How does this force compare to the gravitational force an electron would experience at sea level?
-n~
1V"a. vtt tt.tt ()" $\~ I
~:::. ~ 't
f
I
L 1-1 2..
is a solution to Maxwell's wave equa-
3. Show that the function E = E 0 e-i(kzz-wr)
tion (eqn. 2.26), assuming that c = 1/(µ 0 .s 0 ) 112 .
4. Consider two waves, having slightly different angular frequencies w 1 and cv 2 and
wave numbers k 1 and k 2 , of the form
Ei
where Ei°_ =
be written as
Ef.
=
Ei°e-i (k1z- w 1t)
and
E
2
=
Ege-i (kH. - w 2 t),
Assume their "difference" wave numbers and frequencies can
6k
=
k1 - k2
and
6w
= W1
- w2.
Show that the sum of these two waves leads to a slowly varying amplitude function associated with the difference frequencies and wave numbers multiplied by a
term associated with the actual frequencies and wave numbers. Hint: Reduce the
sum to the product of two cosine functions.
0
-.A (... ~ - w' ~)
E, -::. E1e.
,,h_ I
-:-_
,(..,_
~
'2-
-t"
I.)
-
I~ .__
v-'l- vv
AW
.
~
,'
-
"2_
i; e - j ( ~ ~ -W ~)
l.-o.S [{ ( t. ~ °i- - ./.\. u.J ~
\..__-~--. - . A•~-'-
5{~w'1 vo.v-~1'.-i7 a~p/,'frtde
c f..t z.
5. Show that the group velocity and the phase velocity are equal when there is no
dispersion and are different when dispersion is present. Hint: Begin with the def-
initions of group and phase velocity.
Vp
ol ~
cut
- -
~
vb~
-
w
~
~
w
-4<.. = Vp -
I
~w
n- d'%w- V\..
c..
cl
-C./rt
uJ -
........,.,.
..+
~•Ollil"'1M
Y\...
-w eldW
l
Y\
~W-=. D
Vcr = %_-=- Vp
I-+
d; s p e.f"-S ; ~ "I.
Tl._e.
VI.
a~~ Vb- -:/ Vr;;,
~~
*
0
V\_W
c_
6. Obtain (2.101) from (2.90) and (2.95).
slA k s:t,· r~ n117 : .
2
(j /1) ~
11_'\
/:
(1 + ~e~ b~6~-~~-/~wJ)
....... ]
[ w ~-·'- \ .... ........-"w
I +
b
l
5j
• -- Y-
w""
.
- ,I
:net•-~ ·- '·~~
r
7. Express (2.102) and (2.103) in terms of frequency (instead of angular frequency)
and r (instead of y).
f l ) --
vv
27/ v
-~
-/
--,)
r -- Jd_
'21/-
8. For a medium (such as a gas) of low density and with a single resonance frequency,
show that , if the imaginary part K of the complex index of refraction is significantly smaller than the real part TJ, then for values of w far from w 0 the expressions
for TJ and K can be expressed as follows:
2
TJ
= 1+
Ne
2mso
(
?
1
w0-
)
w2
.
and
2
K
=
Ne
2mso
k
V\_
"2
S'
- [1
f""\J
V\_~
+ Vit/\le4l
)
(w6 - w 2 ) 2
"L
(., {.) - w j)
>> Cl
)( <_ <_ lU
J
2,
~ "'
V\)_ -
yw
.
(
.
rv-
:= w
'L
6pt1 ~_,.j f Y4-1l/41fC!es
I+
(
I
w,}· -
\l t/~
uiU
,-.J
(,,l ~ ,,,
f 'f'
t)
10
2J
-/I)
:r. y ,/
l'lft.tl
V\
~ ~ bo t.rt..
L# 2
9. Assume that a dielectric material has a single resonance frequency at v 0
3 x 10 14 Hz, the polarization decay time is r = 2 x 10-7 s, and the density
of polariiable charges is 5 x 10 26 m- 3 . Determine the full width at half maximum
(FWHM) of the ~bsorption resonance in the material and determine the maximum
numerical value of a.
ex. -=.
"2.
w
c.
I<..
(),_ ~e-1
c, '2.
-($'-w '-
c
Lf
\c..l-=. - I
+~
2
~ rw""'i
=!
l..
V\ \< :: .fJ_
¥W
~>.,_w .._
'-/ ( I + I<.'°) l<=-l"'
o v-
le.
tf
+--
2
/t::. -
c,
'-/
l
·-
I + l
- °2 - l.
-
r 7 w 'l.
~\+
0
__.
C1
'4..
'll'J...W .._
10 (Uivit)
2.2
2
1.8
1.'6
Y\
1.4
1.2
1
I
'0.8
. I
0.6
0-4
0.2
0.92
0.94
0.96
0.98
1
1.02
1.04
':1:()6
1.0:8
1.1
W/w/)
1.8
1.6
1.4
1.2
K
1
0.8
0.6
0.4
/
0.2
01.--~-1-~-d:=====:i::::::....~_L_~-"'~-~-======-~.l--~....L~_J
0.92
0.94
0.%
0.98
W/U/c
1
1.02
1.04
1.06
1.08
1.1
LH 2..
10. A species of atomic weight 60 is doped into an Ah 0 3 crystal at a concentration of
0.1 o/o by weight (cf. Section 5.3). The combined material is found to have an absorbing feature that peaks at 750 nm and a damping constant of 10 13 s- 1. Assume
that each atom of the species contributes to the macroscopic polarization associated with that absorbing feature . Make a plot of TJ and K versus wavelength in the
wavelength region of the absorbing feature.
)... =- 7
¥
=:
_s--~ ~
IM..
It/~
- lli ~ 3,Cf9xtoi>~-~ XI~ -
,,!.\'-
c, - ~~ - "' >tur.J1 ~. ~s-xio- 1 '
::
j,£.7)(10 ~:
-
t if_xto_
l'}
. .,
{_/-1 2.
11. Show that, for a low-density medium such as a gas (as in Problem 8), the maximum and minimum values of r; are located at frequencies that correspond to the
half-maximum values of K. Hint: Assume that in this frequency region y 2 w 2 >>
Cw6 - w2 ) 2 near resonance.
'
(_'<""\ + A k)
~ = I +- Lis:.?
(
V1A t-Q
~3
. F & v- ~ 1tA.. $
t;t
1-1-t."-tO. \
c.i\..-J..
N ~ I /) Y~ l
W
\/'\ ·+.)
r
~
/
I
II""
-....
Al,J!l..
~· rv 3XI tJ
'=?
(!- rr X I D1'I/' \ '2.. ~
t./ X I D
2.~
2. .;:!/
1
:::::
-
•
exptt~S I ()"\
l .+
.l.
'2...
t::.J. ~,..~
1-"\et
~()
.for
t
,,,,...d x.
r, ' x )11'- --;"\,,,. I +- -k x +- · · •·
L'
I
\_
~
-r
W0
I ..t:
A.
-
W
' v
-A O
\
WJ
=I + i ~e~ [Lw0 -w)~;fw)::::i d'~
,
I
I
'2 """-'
\ ~
\
w ~ -..., 'r w)
>'? ~ ~: ( w. •- ~ "..'. ::rvw}
"
"'T":
. (;\. ~ I"'"{
' "-·ti t0 f'
• '•
W/ -
12. In regions of the spectrum where the absorption is small, TJ can be expressed as
TJ2
= 1+
Ne2 (
w6 - w2
)
mco Cw6 - w2)2 + y2w2 .
Convert this equation to wavelength A. instead of angular frequency w. Show that
it is of the form
Use the following index data for crown glass in the expression just displayed to
plot (TJ 2 - 1)- 1 versus A.- 2 and obtain the values of C, N, and A. 0 .
L..
Wavelength
(nm)
Index of
refraction
728.135
706.519
1.5346
1.5352
667.815
587.562
504.774
501.567
492.193
471.314
447.148
438.793
414.376
412.086
401.619
388.865
1.53629
1.53954
1.54417
1.54473
1.54528
1.54624
1.54943
1.55026
1.55374
1.55402
1.55530..
1.55767
Y\ :
Y; J
+
t
v--l~ ( '1'-\.
l,lo\
(!3!,/- ~
..,., ..
I"-"""'-
\r\
i..
1 ~
Ne..... [
~eo
'
W
t\,
- \ ";:: -
'I..) 2..
W
tvu '-- w L...
~,,, "l_ w'l...)"+
~-
>'/ '(
1
Ne.). [
•
M.l. t;)
CN'cl
-
161
hS. ~ v--p1ttn1 I-~
0..
W L.
I
't -
:J
.. -- .
L-tJ "a...
.
u ~ clct-t'"t 1'b
C
(_-::: ~ &ol7;1fC.)
= 7.7f;
f.- 0
N
-==
':>
\....
IV e ..._
o i-ira..~lo\.
C) N > 1\ u
x 10-'~11'\Al..
7
$
~~ a...~!.t.tiC4.
- - Ne" f
\ . .. 1
- ~l~~~)2.-~~!:)LJ
w~
1
'('w'-j
ltOIS-x.10- V1'A.
/,'-('( X / 02."retfa*$/'M'J
~t.t.
uJ )Wo
C. # 2.
13. Two electromagnetic waves in the blue spectral region (450 nm) are separated in
frequency by 1 MHz and in wavelength by 1 pm. What is the group velocity of ·
this wave combination?
W= 27TV
- -
Rl'Tl ...~
.
.
-----c_ II ~
14. Two electromagnetic waves of nearly the same (600-nm) wavelength are measured on a spectrum analyzer to have a frequency difference of 1.0 x 10 8 Hz.
While traveling through a dispersive medium, an interferometer is used to mea- ·_
sure their w.avelength difference of 0.0010 nm. What is the group velocity of the
wave combination in that medium?
A: :
A.
~ 00 f\ V\,;\
v -.:.
\I
'".:
V&
v& :
ltJ ~
Ht
A W
-
Ak. )_ '-
Ak=
'L1T A /..)
~k
~
AA
_
3J['
A'l-
AA
((,., OOXll.l-"1,,._f /O
.
%
{)r 1JC/ XI~_.,,.,;;0
c. fl
15. Determine what emission frequency width would be required to have a temporal
coherence length of 10 m at a source wavelength of 488 nm.
)\-;;. 1..-t
q, Cl
V\ 11\A.._
>-I..._
.ci,\
/0
~
2-
CHL
16. If a photographic film has a minimum resolution of 10 µ.,m, what minimum feature
size could be observed at a distance of 2 m without observing coherent effects?
(Assume the minimum feature size is equal to the minimum resolution.)
a. s. s t.it. "'~
A :::.
S" oo
y-::: "L
w-.,.
"'- !;I.A